# Answer to Question #6439 in Algebra for anil jha

Question #6439

minimum of a function

In the following roblem,

find the least value of the function:

f(x)= +|x-a| +|x-b| + |x-c|

where x is real and, a<B<C< real fixed are a,b,c,d>

i tried using the AM-GM inequality but it was of no use. how should i start?

In the following roblem,

find the least value of the function:

f(x)= +|x-a| +|x-b| + |x-c|

where x is real and, a<B<C< real fixed are a,b,c,d>

i tried using the AM-GM inequality but it was of no use. how should i start?

Expert's answer

Notice that f(x) is the sum of three non-negative functions

|x-a|,

|x-b|, |x-c|

having minimums at a,b,c respectively.

Also notice that

since a<b<c

f(a) = |a-a| + |a-b| + |a-c| = b-a+c-a =

b+c-2a

f(b) = |b-a| + |b-b| + |b-c| = b-a+c-b = c-a

f(c) = |c-a| +

|c-b| + |c-c| = c-a+c-b = 2c-a-b

Since

f(a) = b-a + c-a = b-a +

f(b)

f(c) = c-a + c-b = f(b) + c-b

Thus f(b) < f(a), and f(b)

< f(c).

We claim that x=b is the minimum of f, so

min f = f(b) =

c-a.

It remains to show that for any number x distinct from a,b,c we have

that

f(x) > f(b) = c-a.

Consider the following 4

cases:

1) If x<a<b<c, then

|x-c| > c-a, whence

f(x)= |x-a| +|x-b| + |x-c| > c-a = f(b)

2) If

a<x<b<c, then

|x-a| + |x-c| = c-a,

and

|x-b|>0,

whence

f(x)= |x-a| +|x-b| + |x-c| > c-a = f(b)

3) If a<b<x<c, then

|x-a| + |x-c| = c-a,

and

|x-b|>0,

whence

f(x)= |x-a| +|x-b| + |x-c| > c-a = f(b)

4) If a<b<c<x, then

|x-a| > c-a,

whence

f(x)= |x-a| +|x-b| + |x-c| > c-a = f(b)

Thus

min(f) =

f(b) = c-a.

|x-a|,

|x-b|, |x-c|

having minimums at a,b,c respectively.

Also notice that

since a<b<c

f(a) = |a-a| + |a-b| + |a-c| = b-a+c-a =

b+c-2a

f(b) = |b-a| + |b-b| + |b-c| = b-a+c-b = c-a

f(c) = |c-a| +

|c-b| + |c-c| = c-a+c-b = 2c-a-b

Since

f(a) = b-a + c-a = b-a +

f(b)

f(c) = c-a + c-b = f(b) + c-b

Thus f(b) < f(a), and f(b)

< f(c).

We claim that x=b is the minimum of f, so

min f = f(b) =

c-a.

It remains to show that for any number x distinct from a,b,c we have

that

f(x) > f(b) = c-a.

Consider the following 4

cases:

1) If x<a<b<c, then

|x-c| > c-a, whence

f(x)= |x-a| +|x-b| + |x-c| > c-a = f(b)

2) If

a<x<b<c, then

|x-a| + |x-c| = c-a,

and

|x-b|>0,

whence

f(x)= |x-a| +|x-b| + |x-c| > c-a = f(b)

3) If a<b<x<c, then

|x-a| + |x-c| = c-a,

and

|x-b|>0,

whence

f(x)= |x-a| +|x-b| + |x-c| > c-a = f(b)

4) If a<b<c<x, then

|x-a| > c-a,

whence

f(x)= |x-a| +|x-b| + |x-c| > c-a = f(b)

Thus

min(f) =

f(b) = c-a.

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